IODS course project
About the project
I’m a PhD student in Genetics/proteonomics. I’m taking this course to refresh my R skills and learn some new stuff.
R Exercise 2 Analysis
Task 1
Data wrangling exersize is in the data folder.
This dataset contains information about students and their attitude/strategy towards learning and their exam scores.
setwd("~/GitHub/IODS-project/data")
data <- read.table("learning2014.csv", header = TRUE, sep = ",")
head(data)## Gender Age Attitude Points Deep Surf Stra
## 1 F 53 37 25 3.583333 2.583333 3.375
## 2 M 55 31 12 2.916667 3.166667 2.750
## 3 F 49 25 24 3.500000 2.250000 3.625
## 4 M 53 35 10 3.500000 2.250000 3.125
## 5 M 49 37 22 3.666667 2.833333 3.625
## 6 F 38 38 21 4.750000 2.416667 3.625names(data)## [1] "Gender" "Age" "Attitude" "Points" "Deep" "Surf"
## [7] "Stra"dim(data)## [1] 183 7There are 7 variables and 183 people.
Task 2
library(ggplot2)
summary(data)## Gender Age Attitude Points Deep
## F:122 Min. :17.00 Min. :14.00 Min. : 0.00 Min. :1.583
## M: 61 1st Qu.:21.00 1st Qu.:26.00 1st Qu.:18.00 1st Qu.:3.333
## Median :22.00 Median :32.00 Median :22.00 Median :3.667
## Mean :25.58 Mean :31.21 Mean :20.61 Mean :3.696
## 3rd Qu.:27.00 3rd Qu.:37.00 3rd Qu.:26.00 3rd Qu.:4.083
## Max. :55.00 Max. :50.00 Max. :33.00 Max. :4.917
## Surf Stra
## Min. :1.583 Min. :1.250
## 1st Qu.:2.417 1st Qu.:2.562
## Median :2.833 Median :3.125
## Mean :2.792 Mean :3.085
## 3rd Qu.:3.167 3rd Qu.:3.625
## Max. :4.333 Max. :5.000Summary of all variables. Dataset is about students (Age, Gender) and their attitudes towards learning (Attitude, Deep, Stra, Surf) and their combined scores (Points). Combination stats like Deep/Surf/Stra combine several questions from the questionairre togetger.
Deep is a combination of questionaire questions that reflect “Deep Approach” to learning. (Seeking Meaning, Relating Ideas, Use of Evidence)
Surf is a “Surface Approach” to learning. (Lack of Purpose, Unrelated Memorizing, Syllabus-boundness)
Stra is a “Strategic Approach” to learning. (Organized Studying, Time Management)
I will use the graphical library ggplot to do graphics.
#Pie chart of gender
ggplot(data, aes(x=factor(1), fill = factor(Gender)))+geom_bar(width=1)+coord_polar(theta="y")+theme_void()+labs(title="Gender")
There are 122 females and 61 males.
#Histogram of Age
qplot(data$Age, geom="histogram")+labs(title="Histogram for Age") +
labs(x="Age", y="Count")## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
Age distribution is slanted, there are a lot more people in their 20’s than other people. Age groups cannot really be compared directly because the amount of people are so different.
#Boxplot of gender and points
ggplot(data, aes(x=Gender, y=Points, fill=Gender)) +
geom_boxplot()+labs(title="Gender and amount of points")
It looks like Males have slightly more points by average than females (black bar in the middle is average), but the difference might not be significant. Also there’s a lot more females (122) than males so that might affect this.
Points and Attitude:
#Linear regression of Attitude and Points
ggplot(data, aes(x = Attitude, y = Points)) + geom_point() + geom_smooth(method="lm")+labs(title="Linear regression of attitude and points")
cor.test(data$Attitude, data$Points, method="pearson")##
## Pearson's product-moment correlation
##
## data: data$Attitude and data$Points
## t = 4.8513, df = 181, p-value = 2.635e-06
## alternative hypothesis: true correlation is not equal to 0
## 95 percent confidence interval:
## 0.2042082 0.4615619
## sample estimates:
## cor
## 0.3392167Test of simple linear regression. Pearson correlation is significant and low positive, 0.339. (Pearson correlation assumes normalcy, not sure if it applies here)
Scatter plot matrix:
library(GGally)
p <- ggpairs(data, mapping = aes(col=Gender, alpha=0.3), lower = list(combo = wrap("facethist", bins = 20)))
p
Compilation of all sorts of different statistics.
Task 3 & 4
Task 3
Target: Points
linear <- lm(Points ~ Gender+Age+Attitude, data=data)
summary(linear)##
## Call:
## lm(formula = Points ~ Gender + Age + Attitude, data = data)
##
## Residuals:
## Min 1Q Median 3Q Max
## -25.153 -2.520 1.716 5.411 13.022
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 10.23020 3.39239 3.016 0.00294 **
## GenderM -0.39852 1.35699 -0.294 0.76934
## Age -0.08766 0.07944 -1.103 0.27133
## Attitude 0.40862 0.08707 4.693 5.33e-06 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 8.196 on 179 degrees of freedom
## Multiple R-squared: 0.1218, Adjusted R-squared: 0.1071
## F-statistic: 8.278 on 3 and 179 DF, p-value: 3.465e-05This correlation is not significant for gender and age, but is for attitude. I will keep Attitude and remove Gender, Age.
summary(lm(Points~Attitude+Stra+Surf, data=data))##
## Call:
## lm(formula = Points ~ Attitude + Stra + Surf, data = data)
##
## Residuals:
## Min 1Q Median 3Q Max
## -25.151 -3.212 2.233 5.257 13.694
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 4.77806 5.15767 0.926 0.3555
## Attitude 0.37570 0.08308 4.522 1.11e-05 ***
## Stra 1.89680 0.78318 2.422 0.0164 *
## Surf -0.62623 1.14945 -0.545 0.5866
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 8.082 on 179 degrees of freedom
## Multiple R-squared: 0.146, Adjusted R-squared: 0.1317
## F-statistic: 10.2 on 3 and 179 DF, p-value: 3.089e-06Linear regression is significant for Attitude and Stra, but not Surf. Remove Surf.
Task 4
summary(lm(Points~Attitude+Stra+Deep, data=data))##
## Call:
## lm(formula = Points ~ Attitude + Stra + Deep, data = data)
##
## Residuals:
## Min 1Q Median 3Q Max
## -23.499 -2.557 1.755 5.097 15.420
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 9.69651 4.84553 2.001 0.04689 *
## Attitude 0.40408 0.08178 4.941 1.78e-06 ***
## Stra 2.07504 0.77411 2.681 0.00804 **
## Deep -2.19240 1.08585 -2.019 0.04497 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 7.998 on 179 degrees of freedom
## Multiple R-squared: 0.1637, Adjusted R-squared: 0.1496
## F-statistic: 11.68 on 3 and 179 DF, p-value: 5.036e-07Attitude, Strategy, and trying to get a Deep understanding of the subject during studying explains exam results the best. Here multiple linear regression and all subregressions are significant. So best predictors of a good exam score is the overall attitude of the student and study habits that aim at: Seeking Meaning, Relating Ideas, Use of Evidence, Organized Studying, and Time Management.
R-squared is a statistical measure of how close the data are to the fitted regression line. If the data formed linearly on the regression line, R-squared would be 1 or close to 1. Multiple R-squared is low and that says that these variables do not explain the whole variation in exam points. There are other factors that affect the distribution/orientation of the data.
ggplot(data, aes(x=Points, y=Attitude+Stra+Deep))+geom_point()+geom_smooth(method="lm")+labs(title="Linear regression of points and attitude+stra+deep")
In the picture there’s slight positive linear correlation between Points and Attitude+Stra+Deep, meaning higher scores for an individual in Attitude+Stra+Deep generally mean there’s also a higher Points. While Points correlates with Attitude+Stra+Deep, those variables do not explain all variation in Points. Other factors that could affect points might be how familiar the student is with the course material previously and how much absolute time they spent on studying.
Task 5
linear <- lm(Points~Attitude+Stra+Deep, data=data)
par(mfrow = c(2,2))
plot(linear, which = c(1,2,5))
Q-Q Plot shows that the left tail of the distribution does not follow normalcy. It deviates a lot. I assume left tail contains all the “0 points” people, who answered the questions but possibly didn’t show up for the exam (0 points.) These people should be in a real analysis of the data.
There is also high spread in residuals, but they look mostly grouped linearly.
There is one higher leverage outlier in the data at 0.10 Leverage, while others are mostly between 0.0-0.04.
Chapter 3
Task 1-2
setwd("~/GitHub/IODS-project/data")
alc <- read.table("~/GitHub/IODS-project/data/alc.csv", header = TRUE, sep = ",")
names(alc)## [1] "school" "sex" "age" "address" "famsize"
## [6] "Pstatus" "Medu" "Fedu" "Mjob" "Fjob"
## [11] "reason" "nursery" "internet" "guardian" "traveltime"
## [16] "studytime" "failures" "schoolsup" "famsup" "paid"
## [21] "activities" "higher" "romantic" "famrel" "freetime"
## [26] "goout" "Dalc" "Walc" "health" "absences"
## [31] "G1" "G2" "G3" "alc_use" "high_use"This is a dataset made from two Portuguese school data attributes include descriptors of students (like age, sex, family size, profession of mom and dad, grades etc) and alcohol usage (Dalc = workday alcohol consumption, Walc = weekend alcohol consumption). Alc_use combines Dalc and Walk, and high_use is alcohol usage that’s higher than 2 (1- very low, 5- very high).
Task 3-4
Age
Legal drinking age in Portugal is 16 for beer/cider and 18 for hard liquor. I would hope that there are significantly more high alchohol users in students that are of legal drinking age.
library(ggplot2)
alc$legal <- ifelse(alc$age >= 16, TRUE, FALSE)
ggplot(data = alc, aes(x=high_use))+geom_bar()+facet_wrap("legal")+geom_text(stat='count', aes(label=..count..))
#Students under legal drinking age that are high users
18/(63+18)## [1] 0.2222222#Students over legal drinking age that are high users
96/(205+96)## [1] 0.3189369Here groupings are based on is the student legal drinking age (TRUE/FALSE). The graph shows that there are much more responders that are of legal drinking age. Percentage of students that are high users and under the legal drinking age is 22% and number of students that are high users and over the legal drinking age is 32%. There’s a noticable difference.
Go out
26 goout - going out with friends (numeric: from 1 - very low to 5 - very high)
Assumption: Kids that go out alot with their friends are more likely to drink than those who don’t.
ggplot(data = alc, aes(x=high_use))+geom_bar()+facet_wrap("goout")+geom_text(stat='count', aes(label=..count..))
ftable(table(alc$high_use, alc$goout))## 1 2 3 4 5
##
## FALSE 19 84 103 41 21
## TRUE 3 16 23 40 32Here it’s very obvious, kids that go out a lot drink a lot more than kids that don’t go out much.
Grades
Assumption: Kids who get more than half the points in every grade (First period, second period, final grade) are less likely to drink a lot than kids than don’t even manage half the points.
alc$grades <- ifelse(alc$G1 >= 11 & alc$G2 >= 11 & alc$G3 >=11, TRUE, FALSE)
ggplot(data = alc, aes(x=high_use))+geom_bar()+facet_wrap("grades")+geom_text(stat='count', aes(label=..count..))
#false
66/(66+118)## [1] 0.3586957#true
48/(48+150)## [1] 0.2424242This looks like it might be significant, kids that get more than half the points are 24% high alcohol users and kids that don’t are 35.9% high alchohol users.
Studytime
Assumption: Kids who spend more time studying drink less.
ggplot(data = alc, aes(x=high_use))+geom_bar()+facet_wrap("studytime")+geom_text(stat='count', aes(label=..count..))
ftable(table(alc$high_use, alc$studytime))## 1 2 3 4
##
## FALSE 58 135 52 23
## TRUE 42 60 8 4#1 < 2hours
42/(42+58)## [1] 0.42#2 2 to 5 hours
60/(60+135)## [1] 0.3076923#3 5 to 10 hours
8/(8+52)## [1] 0.1333333#4 over 10 hours
4/(4+23)## [1] 0.1481481Here it’s pretty clear, kids that study more than 5 hours a week are much less likely to be drinkers (14.8%, 13.3%) and kids who study less are much more likely to be heavy drinkers ( 30.7% and 42%). Especially if the kids spend almost no time studying (<2).
Task 5
Age
m <- glm(high_use ~ legal, data = alc, family = "binomial")
summary(m)##
## Call:
## glm(formula = high_use ~ legal, family = "binomial", data = alc)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -0.8765 -0.8765 -0.8765 1.5118 1.7344
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -1.2528 0.2673 -4.687 2.77e-06 ***
## legalTRUE 0.4941 0.2945 1.678 0.0934 .
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 465.68 on 381 degrees of freedom
## Residual deviance: 462.70 on 380 degrees of freedom
## AIC: 466.7
##
## Number of Fisher Scoring iterations: 4coef(m)## (Intercept) legalTRUE
## -1.2527630 0.4941012confint(m)## Waiting for profiling to be done...## 2.5 % 97.5 %
## (Intercept) -1.80548222 -0.7512117
## legalTRUE -0.06548363 1.0947388Age is not significant factor in 95% statistical significance, but would be 90% statistical significance.
Go out
m <- glm(high_use ~ goout, data = alc, family = "binomial")
summary(m)##
## Call:
## glm(formula = high_use ~ goout, family = "binomial", data = alc)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.3721 -0.7672 -0.5450 0.9945 2.3082
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -3.3512 0.4151 -8.073 6.84e-16 ***
## goout 0.7596 0.1157 6.564 5.23e-11 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 465.68 on 381 degrees of freedom
## Residual deviance: 415.68 on 380 degrees of freedom
## AIC: 419.68
##
## Number of Fisher Scoring iterations: 4coef(m)## (Intercept) goout
## -3.3511801 0.7596076confint(m)## Waiting for profiling to be done...## 2.5 % 97.5 %
## (Intercept) -4.1958512 -2.5655359
## goout 0.5385448 0.9931232How often kids go “out” with their friends is significant factor that explains high alcohol use.
Grades
m <- glm(high_use ~ grades, data = alc, family = "binomial")
summary(m)##
## Call:
## glm(formula = high_use ~ grades, family = "binomial", data = alc)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -0.9426 -0.9426 -0.7452 1.4320 1.6835
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -0.5810 0.1537 -3.78 0.000157 ***
## gradesTRUE -0.5584 0.2261 -2.47 0.013526 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 465.68 on 381 degrees of freedom
## Residual deviance: 459.51 on 380 degrees of freedom
## AIC: 463.51
##
## Number of Fisher Scoring iterations: 4coef(m)## (Intercept) gradesTRUE
## -0.5810299 -0.5584044confint(m)## Waiting for profiling to be done...## 2.5 % 97.5 %
## (Intercept) -0.8871701 -0.2834717
## gradesTRUE -1.0051883 -0.1174594Bad grades mean kid is more likely to be a high alcohol user and higher grades mean low chance.
Studytime
m <- glm(high_use ~ studytime, data = alc, family = "binomial")
summary(m)##
## Call:
## glm(formula = high_use ~ studytime, family = "binomial", data = alc)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.0603 -0.8314 -0.8314 1.2993 2.1010
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 0.3209 0.3076 1.043 0.297
## studytime -0.6029 0.1530 -3.941 8.1e-05 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 465.68 on 381 degrees of freedom
## Residual deviance: 448.31 on 380 degrees of freedom
## AIC: 452.31
##
## Number of Fisher Scoring iterations: 4coef(m)## (Intercept) studytime
## 0.3209036 -0.6028561confint(m)## Waiting for profiling to be done...## 2.5 % 97.5 %
## (Intercept) -0.2772274 0.9308876
## studytime -0.9123998 -0.3115583If a kid spends a lot tipe studying they are less likely to be high alchohol users.
Exersize 6
Go out, Grades, STudytime
library(dplyr)##
## Attaching package: 'dplyr'## The following object is masked from 'package:GGally':
##
## nasa## The following objects are masked from 'package:stats':
##
## filter, lag## The following objects are masked from 'package:base':
##
## intersect, setdiff, setequal, unionm <- glm(high_use ~ studytime+grades+goout, data = alc, family = "binomial")
summary(m)##
## Call:
## glm(formula = high_use ~ studytime + grades + goout, family = "binomial",
## data = alc)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.6683 -0.8015 -0.5478 0.9613 2.5753
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -1.9907 0.5184 -3.840 0.000123 ***
## studytime -0.5737 0.1659 -3.459 0.000543 ***
## gradesTRUE -0.3014 0.2485 -1.213 0.225280
## goout 0.7340 0.1171 6.269 3.64e-10 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 465.68 on 381 degrees of freedom
## Residual deviance: 399.34 on 378 degrees of freedom
## AIC: 407.34
##
## Number of Fisher Scoring iterations: 4coef(m)## (Intercept) studytime gradesTRUE goout
## -1.9906826 -0.5736953 -0.3013601 0.7340343confint(m)## Waiting for profiling to be done...## 2.5 % 97.5 %
## (Intercept) -3.0311038 -0.9938476
## studytime -0.9097114 -0.2575669
## gradesTRUE -0.7897173 0.1864329
## goout 0.5103598 0.9703992p <- predict(m, type="response")
# add the predicted probabilities to 'alc'
alc <- mutate(alc, probability = p)
# use the probabilities to make a prediction of high_use
alc <- mutate(alc, prediction = probability > 0.5)
table(high_use = alc$high_use, prediction = alc$prediction) %>% prop.table() %>% addmargins()## prediction
## high_use FALSE TRUE Sum
## FALSE 0.64659686 0.05497382 0.70157068
## TRUE 0.19109948 0.10732984 0.29842932
## Sum 0.83769634 0.16230366 1.00000000g <- ggplot(alc, aes(x = high_use, y = probability, col=prediction))
g + geom_point()
# define a loss function (mean prediction error)
loss_func <- function(class, prob) {
n_wrong <- abs(class - prob) > 0.5
mean(n_wrong)
}
# call loss_func to compute the average number of wrong predictions in the (training) data
loss_func(class = alc$high_use, prob = alc$probability)## [1] 0.2460733There’s a 25% chance of a wrong prediction, so I don’t think my variables are a excellent predictor. It would predict “mostly”, but there’s such a big chance of error it’s not good/usable. Also there’s probably a lot of “overlap” between go out and studytime and studytime and grades… Different predictors would have maybe captured more of the actual problem?
Exersize 7
library(boot)
cv <- cv.glm(data = alc, cost = loss_func, glmfit = m, K = 10)
# average number of wrong predictions in the cross validation
cv$delta[1]## [1] 0.2513089Datacamp model was 0.26 and mine is 0.25… SO they are pretty much the same?
Trying to get a better model:
m <- glm(high_use ~ studytime+failures+goout+absences+sex, data = alc, family = "binomial")
summary(m)##
## Call:
## glm(formula = high_use ~ studytime + failures + goout + absences +
## sex, family = "binomial", data = alc)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -2.0668 -0.7736 -0.5080 0.7663 2.4984
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -3.27654 0.59805 -5.479 4.28e-08 ***
## studytime -0.38032 0.17414 -2.184 0.028966 *
## failures 0.24195 0.21713 1.114 0.265141
## goout 0.71219 0.12037 5.917 3.28e-09 ***
## absences 0.07748 0.02242 3.456 0.000548 ***
## sexM 0.76991 0.26620 2.892 0.003825 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 465.68 on 381 degrees of freedom
## Residual deviance: 380.06 on 376 degrees of freedom
## AIC: 392.06
##
## Number of Fisher Scoring iterations: 4coef(m)## (Intercept) studytime failures goout absences sexM
## -3.27653834 -0.38032241 0.24195066 0.71218501 0.07747656 0.76991496confint(m)## Waiting for profiling to be done...## 2.5 % 97.5 %
## (Intercept) -4.48345712 -2.13293535
## studytime -0.73128369 -0.04615332
## failures -0.18407800 0.67190587
## goout 0.48205204 0.95506124
## absences 0.03450819 0.12366371
## sexM 0.25194047 1.29791469p <- predict(m, type="response")
# add the predicted probabilities to 'alc'
alc <- mutate(alc, probability = p)
# use the probabilities to make a prediction of high_use
alc <- mutate(alc, prediction = probability > 0.5)
table(high_use = alc$high_use, prediction = alc$prediction) %>% prop.table() %>% addmargins()## prediction
## high_use FALSE TRUE Sum
## FALSE 0.65968586 0.04188482 0.70157068
## TRUE 0.17015707 0.12827225 0.29842932
## Sum 0.82984293 0.17015707 1.00000000g <- ggplot(alc, aes(x = high_use, y = probability, col=prediction))
g + geom_point()
# define a loss function (mean prediction error)
loss_func <- function(class, prob) {
n_wrong <- abs(class - prob) > 0.5
mean(n_wrong)
}
# call loss_func to compute the average number of wrong predictions in the (training) data
loss_func(class = alc$high_use, prob = alc$probability)## [1] 0.2120419cv <- cv.glm(data = alc, cost = loss_func, glmfit = m, K = 10)
# average number of wrong predictions in the cross validation
cv$delta[1]## [1] 0.2277487With studytime+failures+goout+absences+sex it drops to 22.5… So maybe that’s good enough!
Chapter 4
Task 1-3
data("Boston")
# explore the dataset
str(Boston)## 'data.frame': 506 obs. of 14 variables:
## $ crim : num 0.00632 0.02731 0.02729 0.03237 0.06905 ...
## $ zn : num 18 0 0 0 0 0 12.5 12.5 12.5 12.5 ...
## $ indus : num 2.31 7.07 7.07 2.18 2.18 2.18 7.87 7.87 7.87 7.87 ...
## $ chas : int 0 0 0 0 0 0 0 0 0 0 ...
## $ nox : num 0.538 0.469 0.469 0.458 0.458 0.458 0.524 0.524 0.524 0.524 ...
## $ rm : num 6.58 6.42 7.18 7 7.15 ...
## $ age : num 65.2 78.9 61.1 45.8 54.2 58.7 66.6 96.1 100 85.9 ...
## $ dis : num 4.09 4.97 4.97 6.06 6.06 ...
## $ rad : int 1 2 2 3 3 3 5 5 5 5 ...
## $ tax : num 296 242 242 222 222 222 311 311 311 311 ...
## $ ptratio: num 15.3 17.8 17.8 18.7 18.7 18.7 15.2 15.2 15.2 15.2 ...
## $ black : num 397 397 393 395 397 ...
## $ lstat : num 4.98 9.14 4.03 2.94 5.33 ...
## $ medv : num 24 21.6 34.7 33.4 36.2 28.7 22.9 27.1 16.5 18.9 ...dim(Boston)## [1] 506 14summary(Boston)## crim zn indus chas
## Min. : 0.00632 Min. : 0.00 Min. : 0.46 Min. :0.00000
## 1st Qu.: 0.08204 1st Qu.: 0.00 1st Qu.: 5.19 1st Qu.:0.00000
## Median : 0.25651 Median : 0.00 Median : 9.69 Median :0.00000
## Mean : 3.61352 Mean : 11.36 Mean :11.14 Mean :0.06917
## 3rd Qu.: 3.67708 3rd Qu.: 12.50 3rd Qu.:18.10 3rd Qu.:0.00000
## Max. :88.97620 Max. :100.00 Max. :27.74 Max. :1.00000
## nox rm age dis
## Min. :0.3850 Min. :3.561 Min. : 2.90 Min. : 1.130
## 1st Qu.:0.4490 1st Qu.:5.886 1st Qu.: 45.02 1st Qu.: 2.100
## Median :0.5380 Median :6.208 Median : 77.50 Median : 3.207
## Mean :0.5547 Mean :6.285 Mean : 68.57 Mean : 3.795
## 3rd Qu.:0.6240 3rd Qu.:6.623 3rd Qu.: 94.08 3rd Qu.: 5.188
## Max. :0.8710 Max. :8.780 Max. :100.00 Max. :12.127
## rad tax ptratio black
## Min. : 1.000 Min. :187.0 Min. :12.60 Min. : 0.32
## 1st Qu.: 4.000 1st Qu.:279.0 1st Qu.:17.40 1st Qu.:375.38
## Median : 5.000 Median :330.0 Median :19.05 Median :391.44
## Mean : 9.549 Mean :408.2 Mean :18.46 Mean :356.67
## 3rd Qu.:24.000 3rd Qu.:666.0 3rd Qu.:20.20 3rd Qu.:396.23
## Max. :24.000 Max. :711.0 Max. :22.00 Max. :396.90
## lstat medv
## Min. : 1.73 Min. : 5.00
## 1st Qu.: 6.95 1st Qu.:17.02
## Median :11.36 Median :21.20
## Mean :12.65 Mean :22.53
## 3rd Qu.:16.95 3rd Qu.:25.00
## Max. :37.97 Max. :50.00# calculate the correlation matrix and round it
cor_matrix<-cor(Boston) %>% round(digits = 2)
# print the correlation matrix
cor_matrix## crim zn indus chas nox rm age dis rad tax
## crim 1.00 -0.20 0.41 -0.06 0.42 -0.22 0.35 -0.38 0.63 0.58
## zn -0.20 1.00 -0.53 -0.04 -0.52 0.31 -0.57 0.66 -0.31 -0.31
## indus 0.41 -0.53 1.00 0.06 0.76 -0.39 0.64 -0.71 0.60 0.72
## chas -0.06 -0.04 0.06 1.00 0.09 0.09 0.09 -0.10 -0.01 -0.04
## nox 0.42 -0.52 0.76 0.09 1.00 -0.30 0.73 -0.77 0.61 0.67
## rm -0.22 0.31 -0.39 0.09 -0.30 1.00 -0.24 0.21 -0.21 -0.29
## age 0.35 -0.57 0.64 0.09 0.73 -0.24 1.00 -0.75 0.46 0.51
## dis -0.38 0.66 -0.71 -0.10 -0.77 0.21 -0.75 1.00 -0.49 -0.53
## rad 0.63 -0.31 0.60 -0.01 0.61 -0.21 0.46 -0.49 1.00 0.91
## tax 0.58 -0.31 0.72 -0.04 0.67 -0.29 0.51 -0.53 0.91 1.00
## ptratio 0.29 -0.39 0.38 -0.12 0.19 -0.36 0.26 -0.23 0.46 0.46
## black -0.39 0.18 -0.36 0.05 -0.38 0.13 -0.27 0.29 -0.44 -0.44
## lstat 0.46 -0.41 0.60 -0.05 0.59 -0.61 0.60 -0.50 0.49 0.54
## medv -0.39 0.36 -0.48 0.18 -0.43 0.70 -0.38 0.25 -0.38 -0.47
## ptratio black lstat medv
## crim 0.29 -0.39 0.46 -0.39
## zn -0.39 0.18 -0.41 0.36
## indus 0.38 -0.36 0.60 -0.48
## chas -0.12 0.05 -0.05 0.18
## nox 0.19 -0.38 0.59 -0.43
## rm -0.36 0.13 -0.61 0.70
## age 0.26 -0.27 0.60 -0.38
## dis -0.23 0.29 -0.50 0.25
## rad 0.46 -0.44 0.49 -0.38
## tax 0.46 -0.44 0.54 -0.47
## ptratio 1.00 -0.18 0.37 -0.51
## black -0.18 1.00 -0.37 0.33
## lstat 0.37 -0.37 1.00 -0.74
## medv -0.51 0.33 -0.74 1.00# visualize the correlation matrix
corrplot(cor_matrix, method="circle", type="upper", cl.pos="b", tl.pos="d", tl.cex = 0.6)
Boston Dataset describes Housing Values in Suburbs of Boston. Here positive correlations are displayed in blue and negative correlations in red color. Color intensity and the size of the circle are proportional to the correlation coefficients.
f.ex. Age is strongly negatively correlated with weighted mean of distances to five Boston employment centres (dis). And index of accessibility to radial highways (rad) is strongly positively correlated with property tax. Number of rooms is positively corrrelated with median value of homes, whereas lower status of the population is strongly negatively correlated.
Task 4
# center and standardize variables
boston_scaled <- scale(Boston)
# summaries of the scaled variables
summary(boston_scaled)## crim zn indus
## Min. :-0.419367 Min. :-0.48724 Min. :-1.5563
## 1st Qu.:-0.410563 1st Qu.:-0.48724 1st Qu.:-0.8668
## Median :-0.390280 Median :-0.48724 Median :-0.2109
## Mean : 0.000000 Mean : 0.00000 Mean : 0.0000
## 3rd Qu.: 0.007389 3rd Qu.: 0.04872 3rd Qu.: 1.0150
## Max. : 9.924110 Max. : 3.80047 Max. : 2.4202
## chas nox rm age
## Min. :-0.2723 Min. :-1.4644 Min. :-3.8764 Min. :-2.3331
## 1st Qu.:-0.2723 1st Qu.:-0.9121 1st Qu.:-0.5681 1st Qu.:-0.8366
## Median :-0.2723 Median :-0.1441 Median :-0.1084 Median : 0.3171
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.:-0.2723 3rd Qu.: 0.5981 3rd Qu.: 0.4823 3rd Qu.: 0.9059
## Max. : 3.6648 Max. : 2.7296 Max. : 3.5515 Max. : 1.1164
## dis rad tax ptratio
## Min. :-1.2658 Min. :-0.9819 Min. :-1.3127 Min. :-2.7047
## 1st Qu.:-0.8049 1st Qu.:-0.6373 1st Qu.:-0.7668 1st Qu.:-0.4876
## Median :-0.2790 Median :-0.5225 Median :-0.4642 Median : 0.2746
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.6617 3rd Qu.: 1.6596 3rd Qu.: 1.5294 3rd Qu.: 0.8058
## Max. : 3.9566 Max. : 1.6596 Max. : 1.7964 Max. : 1.6372
## black lstat medv
## Min. :-3.9033 Min. :-1.5296 Min. :-1.9063
## 1st Qu.: 0.2049 1st Qu.:-0.7986 1st Qu.:-0.5989
## Median : 0.3808 Median :-0.1811 Median :-0.1449
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.4332 3rd Qu.: 0.6024 3rd Qu.: 0.2683
## Max. : 0.4406 Max. : 3.5453 Max. : 2.9865# class of the boston_scaled object
class(boston_scaled)## [1] "matrix"# change the object to data frame
boston_scaled <- as.data.frame(boston_scaled)
# summary of the scaled crime rate
summary(boston_scaled$crim)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## -0.419367 -0.410563 -0.390280 0.000000 0.007389 9.924110# create a quantile vector of crim and print it
bins <- quantile(boston_scaled$crim)
bins## 0% 25% 50% 75% 100%
## -0.419366929 -0.410563278 -0.390280295 0.007389247 9.924109610# create a categorical variable 'crime'
crime <- cut(boston_scaled$crim, breaks = bins, include.lowest = TRUE, labels = c("low", "med_low", "med_high", "high"))
# look at the table of the new factor crime
table(crime)## crime
## low med_low med_high high
## 127 126 126 127# remove original crim from the dataset
boston_scaled <- dplyr::select(boston_scaled, -crim)
# add the new categorical value to scaled data
boston_scaled <- data.frame(boston_scaled, crime)
# number of rows in the Boston dataset
n <- nrow(boston_scaled)
# choose randomly 80% of the rows
ind <- sample(n, size = n * 0.8)
# create train set
train <- boston_scaled[ind,]
# create test set
test <- boston_scaled[-ind,]
# save the correct classes from test data
correct_classes <- test$crime
# remove the crime variable from test data
test <- dplyr::select(test, -crime)In the scaling we subtract the column means from the corresponding columns and divide the difference with standard deviation so the values become z-scores. This normalizes the data so now it will be normally distributed. For some multivariate techniques such as multidimensional scaling and cluster analysis, the concept of distance between the units in the data is often of considerable interest and importance. When the variables in a multivariate data set are on different scales, it makes more sense to calculate the distances after some form of standardization.
Task 5
# linear discriminant analysis
lda.fit <- lda(crime ~ ., data = train)
# print the lda.fit object
lda.fit## Call:
## lda(crime ~ ., data = train)
##
## Prior probabilities of groups:
## low med_low med_high high
## 0.2623762 0.2425743 0.2425743 0.2524752
##
## Group means:
## zn indus chas nox rm
## low 0.9825152 -0.9301856 -0.12375925 -0.8892654 0.4522815
## med_low -0.1024398 -0.2433300 0.04906687 -0.5403407 -0.1642136
## med_high -0.3629840 0.1462347 0.16959035 0.3180592 0.1697274
## high -0.4872402 1.0171096 -0.04073494 1.0499666 -0.4123478
## age dis rad tax ptratio
## low -0.9143905 0.8749344 -0.6839198 -0.7564057 -0.39430406
## med_low -0.3431249 0.3000535 -0.5564696 -0.4850407 -0.02706526
## med_high 0.3802523 -0.3619787 -0.3748243 -0.3025580 -0.32117645
## high 0.7939443 -0.8561960 1.6382099 1.5141140 0.78087177
## black lstat medv
## low 0.38188437 -0.78370134 0.549519292
## med_low 0.30909320 -0.11953033 0.008970193
## med_high 0.08745433 -0.06803167 0.265262047
## high -0.84154179 0.87594429 -0.623220731
##
## Coefficients of linear discriminants:
## LD1 LD2 LD3
## zn 0.05351322 0.82777610 -0.75282575
## indus 0.02935518 -0.28843358 0.31226907
## chas -0.04586880 -0.08926528 0.13524394
## nox 0.45697957 -0.63888934 -1.35507565
## rm -0.08005939 -0.09836615 -0.17219124
## age 0.26655912 -0.29035923 -0.28472946
## dis 0.01378977 -0.33552896 0.05378444
## rad 2.93439655 0.96883696 -0.06386802
## tax 0.02655176 -0.09749300 0.48017985
## ptratio 0.08537027 0.10060183 -0.13474229
## black -0.13256098 -0.01815116 0.10695170
## lstat 0.23483747 -0.24314046 0.48131455
## medv 0.21457481 -0.36302591 -0.17069365
##
## Proportion of trace:
## LD1 LD2 LD3
## 0.9461 0.0404 0.0135# the function for lda biplot arrows
lda.arrows <- function(x, myscale = 1, arrow_heads = 0.1, color = "orange", tex = 0.75, choices = c(1,2)){
heads <- coef(x)
arrows(x0 = 0, y0 = 0,
x1 = myscale * heads[,choices[1]],
y1 = myscale * heads[,choices[2]], col=color, length = arrow_heads)
text(myscale * heads[,choices], labels = row.names(heads),
cex = tex, col=color, pos=3)
}
# target classes as numeric
classes <- as.numeric(train$crime)
# plot the lda results
plot(lda.fit, dimen = 2, col = classes, pch = classes)
lda.arrows(lda.fit, myscale = 1)
Here Linear Discriminant 1 explains most of the between group variance.
rad: index of accessibility to radial highways is the most significant factor that predicts higher crime rate.
Task 6
# predict classes with test data
lda.pred <- predict(lda.fit, newdata = test)
# cross tabulate the results
table(correct = correct_classes, predicted = lda.pred$class)## predicted
## correct low med_low med_high high
## low 10 9 2 0
## med_low 5 21 2 0
## med_high 0 10 18 0
## high 0 0 0 25This prediction is very good at predicting high crime rates, but worse at predicting med_low or med_high correctly. ’ ##Task 7
data("Boston")
new_boston <- scale(Boston)
# k-means clustering
km <-kmeans(new_boston, centers = 4)
# plot the Boston dataset with clusters
pairs(new_boston, col = km$cluster)
#K-means might produce different results every time, because it randomly assigns the initial cluster centers. The function set.seed() can be used to deal with that.
set.seed(123)
# determine the number of clusters
k_max <- 10
# calculate the total within sum of squares
twcss <- sapply(1:k_max, function(k){kmeans(new_boston, k)$tot.withinss})
# visualize the results
qplot(x = 1:k_max, y = twcss, geom = 'line')
# k-means clustering
km <-kmeans(new_boston, centers = 2)
# plot the Boston dataset with clusters
pairs(new_boston, col = km$cluster)
One way to check optimal amount of clusters is to look at the total within cluster sum of squares (twcss). When twcss drops a lot the optimal number of clusters is found. Here twcss drops around 2. In the LDA there were two “main clusters” one with high crime and some points from med_high and the other larger cluster was the rest of the data points. This k-means clustering on this data also works best with 2 clusters.
Bonuses
km <-kmeans(new_boston, centers = 3)
# linear discriminant analysis
lda.fit <- lda(crime ~ ., data = train)
# print the lda.fit object
lda.fit## Call:
## lda(crime ~ ., data = train)
##
## Prior probabilities of groups:
## low med_low med_high high
## 0.2623762 0.2425743 0.2425743 0.2524752
##
## Group means:
## zn indus chas nox rm
## low 0.9825152 -0.9301856 -0.12375925 -0.8892654 0.4522815
## med_low -0.1024398 -0.2433300 0.04906687 -0.5403407 -0.1642136
## med_high -0.3629840 0.1462347 0.16959035 0.3180592 0.1697274
## high -0.4872402 1.0171096 -0.04073494 1.0499666 -0.4123478
## age dis rad tax ptratio
## low -0.9143905 0.8749344 -0.6839198 -0.7564057 -0.39430406
## med_low -0.3431249 0.3000535 -0.5564696 -0.4850407 -0.02706526
## med_high 0.3802523 -0.3619787 -0.3748243 -0.3025580 -0.32117645
## high 0.7939443 -0.8561960 1.6382099 1.5141140 0.78087177
## black lstat medv
## low 0.38188437 -0.78370134 0.549519292
## med_low 0.30909320 -0.11953033 0.008970193
## med_high 0.08745433 -0.06803167 0.265262047
## high -0.84154179 0.87594429 -0.623220731
##
## Coefficients of linear discriminants:
## LD1 LD2 LD3
## zn 0.05351322 0.82777610 -0.75282575
## indus 0.02935518 -0.28843358 0.31226907
## chas -0.04586880 -0.08926528 0.13524394
## nox 0.45697957 -0.63888934 -1.35507565
## rm -0.08005939 -0.09836615 -0.17219124
## age 0.26655912 -0.29035923 -0.28472946
## dis 0.01378977 -0.33552896 0.05378444
## rad 2.93439655 0.96883696 -0.06386802
## tax 0.02655176 -0.09749300 0.48017985
## ptratio 0.08537027 0.10060183 -0.13474229
## black -0.13256098 -0.01815116 0.10695170
## lstat 0.23483747 -0.24314046 0.48131455
## medv 0.21457481 -0.36302591 -0.17069365
##
## Proportion of trace:
## LD1 LD2 LD3
## 0.9461 0.0404 0.0135# the function for lda biplot arrows
lda.arrows <- function(x, myscale = 1, arrow_heads = 0.1, color = "orange", tex = 0.75, choices = c(1,2)){
heads <- coef(x)
arrows(x0 = 0, y0 = 0,
x1 = myscale * heads[,choices[1]],
y1 = myscale * heads[,choices[2]], col=color, length = arrow_heads)
text(myscale * heads[,choices], labels = row.names(heads),
cex = tex, col=color, pos=3)
}
# target classes as numeric
classes <- as.numeric(km$cluster)
# plot the lda results
plot(lda.fit, dimen = 2, col = classes, pch = classes)
lda.arrows(lda.fit, myscale = 1)
rad is the most influencial linear separator for the clusters. It looks like the k-means clustering with 4 clusters makes clusters where low, med_low, med_high, high etc. mix a lot so the clustering is not perfect.
library(plotly)##
## Attaching package: 'plotly'## The following object is masked from 'package:MASS':
##
## select## The following object is masked from 'package:ggplot2':
##
## last_plot## The following object is masked from 'package:stats':
##
## filter## The following object is masked from 'package:graphics':
##
## layoutmodel_predictors <- dplyr::select(train, -crime)
# check the dimensions
dim(model_predictors)## [1] 404 13dim(lda.fit$scaling)## [1] 13 3# matrix multiplication
matrix_product <- as.matrix(model_predictors) %*% lda.fit$scaling
matrix_product <- as.data.frame(matrix_product)
plot_ly(x = matrix_product$LD1, y = matrix_product$LD2, z = matrix_product$LD3, type= 'scatter3d', mode='markers', color= ~train$crime)#plot_ly(x = matrix_product$LD1, y = matrix_product$LD2, z = matrix_product$LD3, type= 'scatter3d', mode='markers', color=~classes)Can’t get the color with km$cluster to work…
Chapter 5
Task 1
human <- read.table("~/GitHub/IODS-project/data/human.csv", header = TRUE, row.names= 1, sep = ",")
# visualize the 'human_' variables
ggpairs(human)
# compute the correlation matrix and visualize it with corrplot
cor(human) %>% corrplot()
summary(human)## edu2Ratio LabRatio LifeExpectancy EduExp
## Min. :0.1717 Min. :0.1857 Min. :49.00 Min. : 5.40
## 1st Qu.:0.7264 1st Qu.:0.5984 1st Qu.:66.30 1st Qu.:11.25
## Median :0.9375 Median :0.7535 Median :74.20 Median :13.50
## Mean :0.8529 Mean :0.7074 Mean :71.65 Mean :13.18
## 3rd Qu.:0.9968 3rd Qu.:0.8535 3rd Qu.:77.25 3rd Qu.:15.20
## Max. :1.4967 Max. :1.0380 Max. :83.50 Max. :20.20
## GNI MatMortality AdoBirth ParliamentF
## Min. : 581 Min. : 1.0 Min. : 0.60 Min. : 0.00
## 1st Qu.: 4198 1st Qu.: 11.5 1st Qu.: 12.65 1st Qu.:12.40
## Median : 12040 Median : 49.0 Median : 33.60 Median :19.30
## Mean : 17628 Mean : 149.1 Mean : 47.16 Mean :20.91
## 3rd Qu.: 24512 3rd Qu.: 190.0 3rd Qu.: 71.95 3rd Qu.:27.95
## Max. :123124 Max. :1100.0 Max. :204.80 Max. :57.50In general female education correlates slightly to moderately positively with “good things” like Life Expectancy and negatively with “bad things” like Maternal Mortality and child pregnancy. Maternal MOrtality in general is very low, but overrepresented in some countries. Maternal Mortality in Europe is generally < 10, but in the some African countries it is very high like Sierra Leone (max, 1100) and Chad (980). Many traits show this same pattern where majority of Western Countries are grouped around the peak and there’s a second smaller “bump” where Developing Countries are. In general most of these trains look good (as in life expectancy is a right tilted distribution, Adolescent birth rate is a left tilted distribution) except parlamentary representation that is left tilted (women are underrepresented in most countries).
Task 2 : Principle Component Analysis
The original variables of data might contain too much information for representing the phenomenon of interest, so we have to reduce dimensions to “most essential dimensions”. We use “Principle component analysis” or PCA, it is an unsupervised method. PCA is sensitive to the relative scaling of the original features and assumes that features with larger viarance are more important that features with smaller variance. Data should be standardized before using PCA.
First Principle Component captures the maximum amount of variance from features in the original data.
Second Principle component is orthogonal to the first and captures maximum amount of variability left.
Not standardized:
#perform principal component analysis (with the SVD method)
pca_human <- prcomp(human)
# draw a biplot
biplot(pca_human, cex = c(0.8, 1), col = c("grey40", "deeppink2"))## Warning in arrows(0, 0, y[, 1L] * 0.8, y[, 2L] * 0.8, col = col[2L], length
## = arrow.len): zero-length arrow is of indeterminate angle and so skipped
## Warning in arrows(0, 0, y[, 1L] * 0.8, y[, 2L] * 0.8, col = col[2L], length
## = arrow.len): zero-length arrow is of indeterminate angle and so skipped
## Warning in arrows(0, 0, y[, 1L] * 0.8, y[, 2L] * 0.8, col = col[2L], length
## = arrow.len): zero-length arrow is of indeterminate angle and so skipped
## Warning in arrows(0, 0, y[, 1L] * 0.8, y[, 2L] * 0.8, col = col[2L], length
## = arrow.len): zero-length arrow is of indeterminate angle and so skipped
## Warning in arrows(0, 0, y[, 1L] * 0.8, y[, 2L] * 0.8, col = col[2L], length
## = arrow.len): zero-length arrow is of indeterminate angle and so skipped
This does not work at all… Standardization is very important. There is not much that can be interpreted here and there’s many errors. Arrows don’t seem to work and the only explaining thing is that GNI explains left grouping and all other variables group right. This is not the same as in corrplot, where GNI has a positive relationship with Life Expectancy and Education Expected Years… This plot doesn’t work.
Task 3 & 4
Standardized
# standardize the variables
human_std <- scale(human)
# perform principal component analysis (with the SVD method)
pca_human <- prcomp(human_std)
# create and print out a summary of pca_human
s <- summary(pca_human)
s## Importance of components:
## PC1 PC2 PC3 PC4 PC5 PC6
## Standard deviation 2.0708 1.1397 0.87505 0.77886 0.66196 0.53631
## Proportion of Variance 0.5361 0.1624 0.09571 0.07583 0.05477 0.03595
## Cumulative Proportion 0.5361 0.6984 0.79413 0.86996 0.92473 0.96069
## PC7 PC8
## Standard deviation 0.45900 0.32224
## Proportion of Variance 0.02634 0.01298
## Cumulative Proportion 0.98702 1.00000# rounded percetanges of variance captured by each PC
pca_pr <- round(100*s$importance[2, ], digits = 1)
# print out the percentages of variance
pca_pr## PC1 PC2 PC3 PC4 PC5 PC6 PC7 PC8
## 53.6 16.2 9.6 7.6 5.5 3.6 2.6 1.3# create object pc_lab to be used as axis labels
pc_lab <- paste0(names(pca_pr), " (", pca_pr, "%)")
# draw a biplot
biplot(pca_human, cex = c(0.8, 1), col = c("grey40", "deeppink2"), xlab = pc_lab[1], ylab = pc_lab[2])
56.1+16.5## [1] 72.6After standardization we see a working plot where we can interpret relationships and groups. Correlations match the correlation seen in corrplot.
Biplot is a way of visualizing two representations of the same data, biplot shows 1) Observations in a lower dimensional representation (scatter plot x and y) and 2) original features and their relationships with each other and the principal components (arrows). Angle of arrows can be interpreted as the correlation between features so that: small angle = high positive correlation and angle between arrow and PC-axis can be interpreted as the correlation between the two (Small angle = high positive correlation). Length of the arrows is proportional to the standard deviation of the features.
Here we see that the principle components explain 72.6% of the variation, so a lot. We also see the same correlations for the corrplot that Maternal Mortality/Adolescent Birthrate group countries towards the right (Developing Countries) and Female Education Indexes and Life expectancy group Countries to the left (European and other Developed Nations). There is strong negative correlation between left and right grouping classes and we can see also that Adolescent Birth Rate nad Maternal mortality correlate closely with eachother… and the same thing is true for the variables to the left. Lab Ratio (Paricipation in Labor male/female ratio) and Parliament (female representation) are in charge of most of the PC2 and only explain 16.5% of the variation. Absolutely clear groups don’t really form and the majority of countries are middle and middle left, only some developing countries crealy separate towards the right.
Task 5 Multiple Correspondance Analysis.
Multiple Correspondance Analysis is also a dimensionality reduction method. It analyses the pattern of relationships between several categorical variables, but can also use continuous variables as supplementary variables. MCA use frequencies and can be used for text data.
For categorical variables: Indicator Matrix (binary) or Burt Matrix (two-way cross tabulatin of all data).
Eigenvalues are variances and percentage of variances retained by each dimension.
Individuals are usually the rows of the data, their coordinates, and contribution to the dimension.
Categories are the coordinates and contribution of the variable categories.
Categorical variables are the squared correlation between each variable and the dimensions.
library(FactoMineR)
library(tidyr)
data(tea)
keep_columns <- c("Tea", "How", "how", "sugar", "where", "lunch")
# select the 'keep_columns' to create a new dataset
tea_time <- select(tea, one_of(keep_columns))
# look at the summaries and structure of the data
summary(tea_time)## Tea How how sugar
## black : 74 alone:195 tea bag :170 No.sugar:155
## Earl Grey:193 lemon: 33 tea bag+unpackaged: 94 sugar :145
## green : 33 milk : 63 unpackaged : 36
## other: 9
## where lunch
## chain store :192 lunch : 44
## chain store+tea shop: 78 Not.lunch:256
## tea shop : 30
## str(tea_time)## 'data.frame': 300 obs. of 6 variables:
## $ Tea : Factor w/ 3 levels "black","Earl Grey",..: 1 1 2 2 2 2 2 1 2 1 ...
## $ How : Factor w/ 4 levels "alone","lemon",..: 1 3 1 1 1 1 1 3 3 1 ...
## $ how : Factor w/ 3 levels "tea bag","tea bag+unpackaged",..: 1 1 1 1 1 1 1 1 2 2 ...
## $ sugar: Factor w/ 2 levels "No.sugar","sugar": 2 1 1 2 1 1 1 1 1 1 ...
## $ where: Factor w/ 3 levels "chain store",..: 1 1 1 1 1 1 1 1 2 2 ...
## $ lunch: Factor w/ 2 levels "lunch","Not.lunch": 2 2 2 2 2 2 2 2 2 2 ...dim(tea_time)## [1] 300 6# visualize the dataset
gather(tea_time) %>% ggplot(aes(value),theme(axis.text.x = element_text(angle = 45, hjust = 1, size = 8))) + facet_wrap("key", scales = "free") + geom_bar()## Warning: attributes are not identical across measure variables;
## they will be dropped
300 tea consumers have answered a survey about their consumption of tea. The questions were about how they consume tea, how they think of tea and descriptive questions (sex, age, socio-professional category and sport practise). Except for the age, all the variables are categorical. The “how” answers to what type of tea people drink and “tea bag” seems most popular, the “How” answers to what people drink ther tea with and most commonly alone. People rarely drink tea at lunch. People put or don’t put sugar in their tea 50/50. Most people drink Earl Grey, and mostly people drink their tea at a chain store.
MCA
# multiple correspondence analysis
mca <- MCA(tea_time, graph = FALSE)
# summary of the model
summary(mca)##
## Call:
## MCA(X = tea_time, graph = FALSE)
##
##
## Eigenvalues
## Dim.1 Dim.2 Dim.3 Dim.4 Dim.5 Dim.6
## Variance 0.279 0.261 0.219 0.189 0.177 0.156
## % of var. 15.238 14.232 11.964 10.333 9.667 8.519
## Cumulative % of var. 15.238 29.471 41.435 51.768 61.434 69.953
## Dim.7 Dim.8 Dim.9 Dim.10 Dim.11
## Variance 0.144 0.141 0.117 0.087 0.062
## % of var. 7.841 7.705 6.392 4.724 3.385
## Cumulative % of var. 77.794 85.500 91.891 96.615 100.000
##
## Individuals (the 10 first)
## Dim.1 ctr cos2 Dim.2 ctr cos2 Dim.3
## 1 | -0.298 0.106 0.086 | -0.328 0.137 0.105 | -0.327
## 2 | -0.237 0.067 0.036 | -0.136 0.024 0.012 | -0.695
## 3 | -0.369 0.162 0.231 | -0.300 0.115 0.153 | -0.202
## 4 | -0.530 0.335 0.460 | -0.318 0.129 0.166 | 0.211
## 5 | -0.369 0.162 0.231 | -0.300 0.115 0.153 | -0.202
## 6 | -0.369 0.162 0.231 | -0.300 0.115 0.153 | -0.202
## 7 | -0.369 0.162 0.231 | -0.300 0.115 0.153 | -0.202
## 8 | -0.237 0.067 0.036 | -0.136 0.024 0.012 | -0.695
## 9 | 0.143 0.024 0.012 | 0.871 0.969 0.435 | -0.067
## 10 | 0.476 0.271 0.140 | 0.687 0.604 0.291 | -0.650
## ctr cos2
## 1 0.163 0.104 |
## 2 0.735 0.314 |
## 3 0.062 0.069 |
## 4 0.068 0.073 |
## 5 0.062 0.069 |
## 6 0.062 0.069 |
## 7 0.062 0.069 |
## 8 0.735 0.314 |
## 9 0.007 0.003 |
## 10 0.643 0.261 |
##
## Categories (the 10 first)
## Dim.1 ctr cos2 v.test Dim.2 ctr
## black | 0.473 3.288 0.073 4.677 | 0.094 0.139
## Earl Grey | -0.264 2.680 0.126 -6.137 | 0.123 0.626
## green | 0.486 1.547 0.029 2.952 | -0.933 6.111
## alone | -0.018 0.012 0.001 -0.418 | -0.262 2.841
## lemon | 0.669 2.938 0.055 4.068 | 0.531 1.979
## milk | -0.337 1.420 0.030 -3.002 | 0.272 0.990
## other | 0.288 0.148 0.003 0.876 | 1.820 6.347
## tea bag | -0.608 12.499 0.483 -12.023 | -0.351 4.459
## tea bag+unpackaged | 0.350 2.289 0.056 4.088 | 1.024 20.968
## unpackaged | 1.958 27.432 0.523 12.499 | -1.015 7.898
## cos2 v.test Dim.3 ctr cos2 v.test
## black 0.003 0.929 | -1.081 21.888 0.382 -10.692 |
## Earl Grey 0.027 2.867 | 0.433 9.160 0.338 10.053 |
## green 0.107 -5.669 | -0.108 0.098 0.001 -0.659 |
## alone 0.127 -6.164 | -0.113 0.627 0.024 -2.655 |
## lemon 0.035 3.226 | 1.329 14.771 0.218 8.081 |
## milk 0.020 2.422 | 0.013 0.003 0.000 0.116 |
## other 0.102 5.534 | -2.524 14.526 0.197 -7.676 |
## tea bag 0.161 -6.941 | -0.065 0.183 0.006 -1.287 |
## tea bag+unpackaged 0.478 11.956 | 0.019 0.009 0.000 0.226 |
## unpackaged 0.141 -6.482 | 0.257 0.602 0.009 1.640 |
##
## Categorical variables (eta2)
## Dim.1 Dim.2 Dim.3
## Tea | 0.126 0.108 0.410 |
## How | 0.076 0.190 0.394 |
## how | 0.708 0.522 0.010 |
## sugar | 0.065 0.001 0.336 |
## where | 0.702 0.681 0.055 |
## lunch | 0.000 0.064 0.111 |# visualize MCA
plot(mca, invisible=c("ind"), habillage="quali")
With the MCA we can see that some things group together. People that drink tea at a tea shop drink unpackaged tea and are more likely to drink green tea than others, where as people that drink tea in a chain store are most likely to drink it in a tea bag. Things that were 50/50 don’t part from the middle of the MCA plot, so sugar/no sugar are in the middle. Also because almost everyone had their drink “Not lunch” it doesn’t group anywhere…
Other plots
plotellipses(mca)
How the chosen columns spread out in the MCA.Here it’s also visible that teashop+unpackaged group together. Also it’s easier to see that it looks like that people that drink unpackaged tea in teashops mostly drink tea alone or with lemon. Less with milk and none or almost none (hard to see) with “other”. Also this group is maybe mostly drinking “not at lunch”.